I was given the following problem:
The probability that a student drives a car, takes a bus, rides a bike or walks to school are 0.4, 0.2, 0.3 and 0.1 respectively. Suppose we randomly select 9 students and let $X_1$ be the number of students that drive to school, $X_2$ be the number that take buses, $X_3$ the number that ride bikes and $X_4$ the number that walk. Let $Y = X_1 + X_2$. Find $E(Y)$ and $Var(Y)$.
I did the following:
$$E(Y) = E(X_1 + X_2) = E(X_1) + E(X_2) = np_1 + np_2 = 9*0.4 + 9*0.2 = 5.4$$
$$Var(Y) = Var(X_1 + X_2) = Var(X_1) + Var(X_2) = np_1(1-p_1) + np_2(1-p_2) = ...=3.6$$
The automated grading software marked the $Var(Y)=3.6$ wrong, but did not provide any insight as to why. Can someone explain this to me?
Thank you!